Difference between revisions of "1997 USAMO Problems/Problem 2"
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<math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. | <math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. | ||
==Solution== | ==Solution== | ||
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Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's Theorem]], The three lines are [[concurrent]] if | Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's Theorem]], The three lines are [[concurrent]] if | ||
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QED | QED | ||
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| + | ==Solution 2== | ||
| + | These three lines concur at the radical center of the three circles centered at <math>D, E, F</math> and passing through <math>C, A, B</math> respectively. (Indeed, each line passes through one intersection point of a pair of circles, and because it is perpendicular to the line of centers it must the the radical axis of these circles.) | ||
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| + | {{alternate solutions}} | ||
==See Also== | ==See Also== | ||
Revision as of 12:21, 13 February 2015
Contents
Problem
is a triangle. Take points 
on the perpendicular bisectors of 
respectively. Show that the lines through 
perpendicular to 
respectively are concurrent.
Solution
Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if
But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.
QED
Solution 2
These three lines concur at the radical center of the three circles centered at and passing through 
respectively. (Indeed, each line passes through one intersection point of a pair of circles, and because it is perpendicular to the line of centers it must the the radical axis of these circles.)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1997 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
